Carbohydrates and the structure of glucose
Link to Std 11 : In Std 11 you learnt about isomerism and chiral carbon atoms; glucose is a standard example of both.
Rice, bread and sugar are mostly carbohydrates, which give the body its main supply of energy.
Carbohydrates are polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis. They are classed as monosaccharides (cannot be hydrolysed), oligosaccharides (2–10 units) and polysaccharides (many units).
In the figure : The boxes show the three classes with examples. Glucose has 4 chiral carbons.
Rule 1: Classification — Monosaccharides: aldose (–CHO) or ketose (C=O) of 3–7 carbons (glucose, fructose, ribose). Disaccharides give two monosaccharides on hydrolysis (sucrose, maltose, lactose). Polysaccharides give many units (starch, glycogen, cellulose). Reducing sugars give Tollens'/Fehling's tests (free –CHO or a free hemiacetal OH).
- Easy : Glucose is an aldohexose; fructose is a ketohexose.
- Medium : Sucrose is a non-reducing disaccharide; maltose and lactose are reducing.
- Exam-type : Molar mass of glucose C₆H₁₂O₆ = 180.16 g mol⁻¹ (C = 12.011, H = 1.008, O = 15.999).
Rule 2: Evidence for the open chain of glucose — Glucose C₆H₁₂O₆ is reduced by HI to n-hexane (straight 6-carbon chain). It gives an oxime with NH₂OH and a cyanohydrin with HCN (a C=O group). Bromine water oxidises it to gluconic acid (–CHO group). Acetylation gives a pentaacetate (five –OH). Nitric acid gives saccharic acid (two –COOH, a –CH₂OH at the end).
- Easy : Glucose + Br₂ water → gluconic acid (confirms –CHO).
- Medium : Glucose + (CH₃CO)₂O → pentaacetate (five –OH groups).
- Exam-type : Glucose + HI (heat) → n-hexane: the carbons form an unbranched chain.
Rule 3: Stereochemistry (D and L) — Glucose has 4 chiral carbons (C-2 to C-5), so 2⁴ = 16 stereoisomers (8 D- and 8 L-forms). The configuration at C-5 (the highest-numbered chiral carbon) decides D (OH on the right in the Fischer projection) or L. Natural glucose is D-(+)-glucose, dextrorotatory; D and L describe the structure, not the direction of optical rotation.
- Easy : D-glucose and L-glucose are enantiomers.
- Medium : The 'D' in D-(+)-glucose refers to the configuration at C-5; (+) means it rotates plane-polarised light to the right.
- Exam-type : A ketohexose such as fructose has 3 chiral carbons, so 2³ = 8 stereoisomers.
Glucose: aldohexose, 4 chiral carbons, 16 stereoisomers; D/L refers to C-5 configuration.
Thinking D means dextrorotatory; D refers to configuration, and (+) refers to rotation.
Memory clue: "D–C5 on the right" — D means the OH on C-5 is on the right.
Spot the difference
| Point | Glucose | Fructose |
|---|---|---|
| Type | Aldohexose | Ketohexose |
| Chiral carbons | 4 | 3 |
| Optical rotation | Dextro (+) | Laevo (−) |
Practice questions
- 1What is a reducing sugar?
- 2How many stereoisomers does glucose have?
- 3Is fructose an aldose or ketose?
- 4Give two pieces of evidence that glucose has an aldehyde group.
- 5What does D in D-glucose mean?
- 6Name the product of glucose oxidised by bromine water.
- 7Explain why glucose gives a pentaacetate and what this shows.
- 8Find the number of stereoisomers of a ketohexose with 3 chiral carbons.
Show answers (try them yourself first)
- 1A sugar that reduces Tollens' or Fehling's reagent.
- 216.
- 3Ketose.
- 4It forms an oxime/cyanohydrin and is oxidised to gluconic acid by bromine water.
- 5The configuration at C-5 matches D-glyceraldehyde (OH on the right).
- 6Gluconic acid.
- 7Five acetyl groups attach because glucose has five –OH groups.
- 82³ = 8.
Cyclic structure, anomers and disaccharides
Link to Std 11 : In Std 11 you learnt that carbonyl compounds react with alcohols (hemiacetal formation); glucose does this within its own molecule.
Table sugar (sucrose) is made from two simpler sugars joined by a link; digesting it breaks the link and gives glucose and fructose.
Glucose exists mainly as a six-membered ring (pyranose) formed by the –OH on C-5 adding to the aldehyde C-1. This creates a new chiral carbon (C-1, the anomeric carbon), giving α and β anomers. Two sugars join by a glycosidic linkage to form disaccharides.
In the figure : α and β differ only at C-1 and interconvert in solution (mutarotation); sucrose has the glycosidic link between C-1 of glucose and C-2 of fructose.
Rule 1: Haworth structure and anomers — In the pyranose ring of D-glucose, C-1 –OH is down in α and up in β (Haworth). The α and β forms have different specific rotations, +112° and +19°. When either is dissolved in water the rotation changes slowly to +52.5° (mutarotation) as an equilibrium mixture forms through the open chain.
- Easy : α-D-glucose: [α] = +112°; β-D-glucose: [α] = +19°; equilibrium = +52.5°.
- Medium : Mole fraction of α at equilibrium: (52.5 − 19) ÷ (112 − 19) = 0.36.
- Exam-type : Why does C-1 become chiral on ring closure? It carries four different groups after hemiacetal formation.
Rule 2: Disaccharides — Sucrose: α-D-glucose (C-1) joined to β-D-fructose (C-2), so both anomeric carbons are involved; it is non-reducing. Hydrolysis gives an equimolar glucose–fructose mixture called invert sugar (sign of rotation changes from + to −). Maltose: two α-D-glucose units (C-1 to C-4); reducing. Lactose: galactose + glucose (β-1,4); reducing.
- Easy : Sucrose → glucose + fructose (H⁺ or invertase).
- Medium : Maltose is reducing because one glucose unit has a free hemiacetal OH.
- Exam-type : Sucrose rotation +66.5° changes to −20° after hydrolysis: the mixture is 'inverted'.
Rule 3: Polysaccharides — Starch: amylose (unbranched, α-1,4, 15–20%) and amylopectin (branched, α-1,4 and α-1,6, 80–85%); the plant's storage form. Glycogen: branched like amylopectin, stored in animals. Cellulose: β-D-glucose units, β-1,4, unbranched, the structural material of plants. Humans digest starch but not cellulose (no cellulase).
- Easy : Starch + iodine → blue colour.
- Medium : Cellulose hydrolysed with acid gives glucose only.
- Exam-type : Why can cows digest cellulose but humans cannot? Cows have bacteria that make the enzyme for β-1,4 links.
α/β differ at C-1 only. Sucrose: non-reducing (both anomeric C used). Starch α-glucose; cellulose β-glucose.
Calling sucrose a reducing sugar; its two anomeric carbons are tied in the glycosidic link.
Memory clue: "Starch α, cellulose β; sucrose cannot reduce".
Spot the difference
| Point | Starch | Cellulose |
|---|---|---|
| Monomer | α-D-glucose | β-D-glucose |
| Role | Energy storage (plants) | Structural (plants) |
| Digestible by humans? | Yes | No |
Practice questions
- 1What are anomers?
- 2Which sugar is called invert sugar?
- 3Which polysaccharide gives a blue colour with iodine?
- 4What is mutarotation?
- 5Why is sucrose non-reducing?
- 6Name the monomer of cellulose and the type of linkage.
- 7Using specific rotations (+112° for α, +19° for β, +52.5° at equilibrium) find the percentage of α-glucose at equilibrium.
- 8Compare amylose, amylopectin and cellulose in structure and function.
Show answers (try them yourself first)
- 1Stereoisomers differing in configuration at the anomeric carbon (C-1).
- 2The mixture of glucose and fructose from sucrose hydrolysis.
- 3Starch.
- 4Gradual change of specific rotation of a fresh glucose solution to an equilibrium value.
- 5Both anomeric carbons are involved in the glycosidic link, so no free hemiacetal group remains.
- 6β-D-glucose, β-1,4 glycosidic linkage.
- 7(52.5 − 19) ÷ (112 − 19) = 0.36, i.e. 36.0 %.
- 8Amylose: unbranched α-1,4 chain; amylopectin: branched α-1,4 and α-1,6; cellulose: β-1,4 unbranched, structural.
Amino acids and proteins
Link to Std 11 : In Std 11 you learnt about –NH₂ and –COOH groups; amino acids carry both, which makes them amphoteric.
Boiling an egg turns the clear white solid because heat unfolds its proteins permanently.
Amino acids contain –NH₂ and –COOH on the α-carbon. They exist as zwitterions (H₃N⁺–CHR–COO⁻) and join by peptide bonds (–CO–NH–) to form polypeptides and proteins. Protein structure has four levels, and heat or acid can denature it.
In the figure : Amino acid → zwitterion → peptide chain → folded protein. Denaturation breaks the folding but not the peptide sequence.
Rule 1: Amino acids and zwitterions — General formula RCH(NH₂)COOH. In solution the acid group gives H⁺ to the amine group, forming a dipolar ion with no net charge, called a zwitterion. They are amphoteric: with acid they form cations, with alkali anions. Glycine (R = H) is the only optically inactive α-amino acid. Essential amino acids must come from food; non-essential are synthesised by the body.
- Easy : Glycine: H₂N–CH₂–COOH (no chiral carbon).
- Medium : In strongly acidic solution an amino acid exists as H₃N⁺–CHR–COOH (cation).
- Exam-type : At the isoelectric point the amino acid is neutral overall and least soluble.
Rule 2: Peptide bond — Two amino acids join with loss of water: –COOH + H₂N– → –CO–NH–. A dipeptide has one peptide bond; a tripeptide two; n amino acids have (n − 1) peptide bonds. Molar mass of a peptide = sum of amino acid masses − 18.02 × (n − 1).
- Easy : Glycine (75.07) + alanine (89.09) → Gly–Ala dipeptide of mass 146.14 g mol⁻¹.
- Medium : A tetrapeptide has 3 peptide bonds.
- Exam-type : Three different amino acids can form 3! = 6 different tripeptides if each is used once.
Rule 3: Levels of protein structure — Primary: sequence of amino acids. Secondary: α-helix or β-pleated sheet held by hydrogen bonds between C=O and N–H. Tertiary: overall 3D folding (disulphide bridges, hydrogen bonds, ionic and hydrophobic interactions). Quaternary: assembly of several polypeptide chains (e.g. haemoglobin, four chains). Fibrous proteins (keratin, myosin) are insoluble; globular proteins (albumin, insulin) are soluble.
- Easy : Keratin in hair: fibrous protein, mainly α-helix.
- Medium : Boiling an egg changes the structure of albumin from a soluble globular form to an insoluble coagulated form.
- Exam-type : Denaturation breaks hydrogen bonds and weak interactions but not the primary structure.
Primary = sequence, 2° = helix/sheet, 3° = folding, 4° = multiple chains. Peptide bond = –CO–NH–.
Thinking denaturation breaks peptide bonds; only the weaker interactions are disrupted.
Memory clue: "1-sequence, 2-helix, 3-fold, 4-teamwork".
Spot the difference
| Point | Fibrous protein | Globular protein |
|---|---|---|
| Shape | Thread-like | Spherical |
| Solubility in water | Insoluble | Soluble |
| Examples | Keratin, myosin | Insulin, albumin |
Practice questions
- 1What bond links amino acids in a protein?
- 2What is a zwitterion?
- 3How many peptide bonds are in a tripeptide?
- 4Why are amino acids amphoteric?
- 5What is denaturation of a protein?
- 6Name the four levels of protein structure.
- 7Find the molar mass of Gly–Ala (glycine 75.07, alanine 89.09, water 18.02).
- 8Explain why boiling does not destroy the nutritional value of a protein's amino acids but does change its function.
Show answers (try them yourself first)
- 1Peptide bond (–CO–NH–).
- 2A dipolar ion carrying both a positive and a negative charge.
- 3Two.
- 4They contain both a basic –NH₂ and an acidic –COOH group.
- 5Loss of secondary and tertiary structure (and function) by heat or chemicals; the primary structure remains.
- 6Primary, secondary, tertiary and quaternary.
- 775.07 + 89.09 − 18.02 = 146.14 g mol⁻¹.
- 8Heat disrupts hydrogen bonds and folding (so function is lost), but the peptide bonds and amino acids stay intact.
Enzymes and vitamins
Link to Std 11 : In Std 11 you learnt about catalysts; enzymes are biological catalysts that act at body temperature.
Curd forms when enzymes made by bacteria act on milk sugar, and a lack of fruit in the diet can lead to bleeding gums because of vitamin C deficiency.
Enzymes are protein catalysts, each specific for a reaction. A vitamin is an organic compound needed in small amounts that the body cannot make enough of. Vitamins are fat-soluble (A, D, E, K) or water-soluble (B group and C).
In the figure : The substrate binds to the enzyme's active site; each vitamin deficiency gives a typical disease.
Rule 1: Enzymes — Enzymes are globular proteins (some need a cofactor). Features: highly specific, work under mild conditions (310 K, pH ≈ 7), speed up reactions by a factor of 10⁶–10¹². The substrate binds to the active site (lock-and-key or induced fit) forming an enzyme–substrate complex. Names end in -ase (maltase, urease). Inhibitors reduce activity; heat or extreme pH denatures the enzyme.
- Easy : Urease: urea + H₂O → 2NH₃ + CO₂.
- Medium : Invertase: sucrose → glucose + fructose.
- Exam-type : Why does a fever above about 313 K harm enzymes? High temperature denatures the protein and changes the active site.
Rule 2: Vitamins and deficiency diseases — A (retinol): night blindness; B₁: beri-beri; B₂: cheilosis; B₆: convulsions; B₁₂: pernicious anaemia; C: scurvy; D: rickets; E: increased fragility of RBC; K: increased blood clotting time. The B group and C are water soluble and are needed daily; A, D, E, K are fat soluble and are stored in the liver and fat tissue.
- Easy : Lack of vitamin C → scurvy; lack of vitamin D → rickets.
- Medium : Vitamin K is needed for blood clotting.
- Exam-type : Why can vitamin A overdose be harmful but vitamin C excess is usually excreted? A is stored in fat; C is water soluble.
Rule 3: Enzyme rate and substrate concentration — Initial rate v = Vmax[S] ÷ (Km + [S]). At [S] = Km the rate is half of Vmax; at very large [S] all active sites are filled and the rate levels off at Vmax.
- Easy : Vmax = 80 µmol min⁻¹ and Km = 2 mM: at [S] = 2 mM, v = 40 µmol min⁻¹.
- Medium : At [S] = 8 mM: v = 80 × 8 ÷ (2 + 8) = 64.0 µmol min⁻¹.
- Exam-type : At [S] = 0.5 mM: v = 80 × 0.5 ÷ 2.5 = 16.0 µmol min⁻¹.
Enzymes are specific protein catalysts; deficiency diseases link a vitamin to a symptom.
Saying enzymes work at any temperature; high temperature denatures them.
Memory clue: "C scurvy, D rickets, A night, B₁ beri-beri".
Spot the difference
| Point | Fat-soluble | Water-soluble |
|---|---|---|
| Vitamins | A, D, E, K | B group, C |
| Storage | Stored in body | Excreted; needed daily |
| Overdose risk | Higher | Lower |
Practice questions
- 1What is an enzyme?
- 2Which disease is caused by lack of vitamin C?
- 3Name two fat-soluble vitamins.
- 4Explain the lock-and-key model.
- 5Why do enzymes lose activity at high temperature?
- 6Which vitamin deficiency causes rickets?
- 7For Vmax = 80 µmol min⁻¹ and Km = 2 mM, find the rate at [S] = 8 mM.
- 8Why must water-soluble vitamins be taken regularly but fat-soluble vitamins need not be?
Show answers (try them yourself first)
- 1A biological catalyst (usually a protein) specific for a reaction.
- 2Scurvy.
- 3Any two of A, D, E, K.
- 4The substrate fits the active site of the enzyme like a key in a lock, forming an enzyme–substrate complex.
- 5Heat denatures the protein and distorts the active site.
- 6Vitamin D deficiency.
- 7v = 80 × 8 ÷ (2 + 8) = 64.0 µmol min⁻¹.
- 8Water-soluble vitamins are excreted in urine; fat-soluble ones are stored in fat tissue.
Nucleic acids and hormones
Link to Std 11 : In Std 11 you learnt about hydrogen bonds; they hold the two strands of DNA together.
Every cell carries instructions in DNA, which is why children resemble their parents.
Nucleic acids (DNA and RNA) are polymers of nucleotides. DNA stores genetic information and RNA carries it out through protein synthesis. Hormones are chemical messengers made by glands and carried by blood.
In the figure : A nucleotide has a base, a sugar and a phosphate; DNA bases pair A–T and G–C.
Rule 1: Nucleotides and nucleic acids — Nucleoside = base + sugar; nucleotide = base + sugar + phosphate. Bases: purines (A, G) and pyrimidines (C, T in DNA; C, U in RNA). DNA has 2-deoxyribose, RNA has ribose. Nucleotides join by phosphodiester links between C-3′ and C-5′ of sugars.
- Easy : DNA bases: adenine, guanine, cytosine, thymine.
- Medium : RNA has uracil in place of thymine.
- Exam-type : A DNA segment is rich in A–T pairs; it melts more easily than a G–C rich segment because A–T has only two hydrogen bonds.
Rule 2: DNA structure and base pairing — DNA is a double helix of two antiparallel strands joined by base pairs: A pairs with T (2 H-bonds), G with C (3 H-bonds). So [A] = [T] and [G] = [C] (Chargaff). Replication gives two identical copies.
- Easy : If a DNA has 30% adenine, it has 30% thymine, and the rest 40% is G + C (20% each).
- Medium : A DNA duplex of 100 base pairs has 100 × 2 = 200 bases in all.
- Exam-type : Hydrogen bond count for 10 A–T and 15 G–C pairs: 10 × 2 + 15 × 3 = 65.
Rule 3: Protein synthesis, hormones — DNA → m-RNA (transcription) → protein (translation at ribosomes with t-RNA). A three-base codon codes for one amino acid: 4³ = 64 possible codons. Hormones: insulin (lowers blood glucose), glucagon (raises it), thyroxine (metabolic rate), adrenaline (emergency), sex hormones (steroids).
- Easy : Codons are triplets: 64 combinations code 20 amino acids (redundant).
- Medium : Insulin deficiency leads to diabetes mellitus.
- Exam-type : A gene of 300 bases codes for at most 300 ÷ 3 = 100 amino acids.
DNA: A–T, G–C, double helix, deoxyribose. RNA: A–U, single strand, ribose.
Pairing A with U in DNA; uracil occurs only in RNA.
Memory clue: "AT-two, GC-three" — A–T has two hydrogen bonds and G–C has three.
Spot the difference
| Point | DNA | RNA |
|---|---|---|
| Sugar | Deoxyribose | Ribose |
| Bases | A, G, C, T | A, G, C, U |
| Strands | Double | Single |
Practice questions
- 1Which base is present in RNA but not DNA?
- 2What does a codon contain?
- 3Name the hormone that lowers blood glucose.
- 4If a DNA sample has 30% adenine, what is the % of guanine?
- 5Why is a G–C pair stronger than an A–T pair?
- 6How many amino acids can a gene of 600 bases at most code for?
- 7Count hydrogen bonds in a duplex with 12 A–T pairs and 20 G–C pairs.
- 8Explain why there are 64 codons but only 20 amino acids.
Show answers (try them yourself first)
- 1Uracil.
- 2Three bases (a triplet).
- 3Insulin.
- 4A = T = 30%, so G + C = 40% and G = 20%.
- 5G–C has three hydrogen bonds; A–T has only two.
- 6At most 600 ÷ 3 = 200 amino acids.
- 712 × 2 + 20 × 3 = 84 hydrogen bonds.
- 8There are 4³ = 64 possible triplets; several codons can code for the same amino acid, so the code is redundant.